INTRODUCTORY NOTES on MODULES 1. Introduction One of the Most Basic Concepts in Linear Algebra Is Linear Combinations

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INTRODUCTORY NOTES on MODULES 1. Introduction One of the Most Basic Concepts in Linear Algebra Is Linear Combinations INTRODUCTORY NOTES ON MODULES KEITH CONRAD 1. Introduction One of the most basic concepts in linear algebra is linear combinations: of vectors, of polynomials, of functions, and so on. For example, the polynomial 7 − 2T + 3T 2 is a linear combination of 1;T , and T 2: 7 · 1 − 2 · T + 3 · T 2. The coefficients used for linear combinations in a vector space are in a field, but there are many places where we meet linear combinations with coefficients in a ring. p p Example 1.1. In the ring Z[ −5] let p be the ideal (2; 1 + −5), so by definition p p p p p p = 2x + (1 + −5)y : x; y 2 Z[ −5] = Z[ −5] · 2 + Z[ −5] · (1 + −5); p p which is the set of all linear combinations of 2 and 1 + −5 with coefficients in Z[ −5]. p Such linear combinations do not provide unique representation: for x and y in Z[ −5], p p p x · 2 + y · (1 + −5) = (x − (1 + −5)) · 2 + (y − 2) · (1 + −5): p So we can't treat x and y as \coordinates" of 2x + (1 + −5)y. In a vector space like Rn or Cn, whenever there is a duplication of representations with a spanning set we can remove a vector from p the spanning set, but in p this is not the case: if we could reduce the spanning set f2; 1 + −5g of p p p to a single element α, then p = Z[ −5]α = (α) would be a principal ideal in Z[ −5], but it can be shown that p is not a principal ideal. p p Instead of reducing the size of f2; 1 + −5g to get a nice spanning set for p with Z[ −5]- coefficients, we can try to do something else to standardize the representation of elements of p as linear combinations: restrict coefficients to Z. That works in this case because: p p p = 2x + (1 + −5)y : x; y 2 Z[ −5] p p p = 2(a + b −5) + (1 + −5)(c + d −5) : a; b; c; d 2 Z p p p = 2a + 2 −5b + (1 + −5)c + (−5 + −5)d : a; b; c; d 2 Z p p p = Z · 2 + Z · 2 −5 + Z · (1 + −5) + Z · (−5 + −5): Describing p in terms of linear combinations with Z-coefficients made the spanning set grow. We p can shrink the spanning set back to f2; 1 + −5g because the new members of this spanning set, p p p 2 −5 and −5 + −5, are Z-linear combinations of 2 and 1 + −5: p p p p 2 −5 = (−1)2 + 2(1 + −5) and − 5 + −5 = (−3)2 + (1 + −5): 1 2 KEITH CONRAD p p p Thus 2 −5 and −5 + −5 are in Z · 2 + Z · (1 + −5), so p p p = Z · 2 + Z · (1 + −5) = 2m + (1 + −5)n : m; n 2 Z : p p Using coefficients in Z rather than in Z[ −5], there is unique representation: if 2m+(1+ −5)n = p 2m0 + (1 + −5)n0 then p (2(m − m0) + (n − n0)) + (n − n0) −5 = 0: The real and imaginary parts are 0, so n = n0 and then m = m0. Thus, we can regard m and n as p \coordinates" for 2m + (1 + −5)n, and the situation looks a lot closer to ordinary linear algebra. p p Warning. That the set f2; 1 + −5g has its Z[ −5]-linear combinations coincide with its Z- p linear combinations is something of a fluke. In the ring Z[ d], the set of Z-linear combinations of p two elements generally does not coincide with the set of their Z[ d]-linear combinations. The use of linear combinations with coefficients coming from a ring rather than a field suggests the concept of \vector space over a ring," which for historical reasons is not called a vector space but instead a module. Definition 1.2. For a commutative ring R, an R-module M is an abelian group M on which R acts by additive maps respecting the ring structure of R when these maps are added and composed: there is a scalar multiplication function R × M ! M denoted by (r; m) 7! rm such that (1)1 m = m for all m 2 M. (2) r(m + m0) = rm + rm0 for all r 2 R and m; m0 2 M. (3)( r + r0)m = rm + r0m and (rr0)m = r(r0m) for all r; r0 2 R and m 2 M. In the special case when R = F is a field, an F -module is just an F -vector space by another name. Example 1.3. The set Rn of ordered n-tuples in R with componentwise addition and scalar multiplication by R as in linear algebra is an R-module. Example 1.4. Every ideal I in R is an R-module with addition and scalar multiplication being the operations in R. Example 1.5. A quotient ring R=I for an ideal I is an R-module where r(x mod I) := rx mod I for r 2 R and x 2 R. (It is easy to check that this is well-defined and satisfies the axioms.) So I and R=I are both R-modules, whereas in the language of ring theory, ideals and quotient rings are not the same kind of object: an ideal is almost never a ring, for instance. Example 1.6. The ring R[T ] is an R-module using obvious addition and scalar multiplication. Example 1.7. The set Map(R; R) of functions f : R ! R under pointwise addition of functions and the scalar multiplication (rf)(x) = r(f(x)) is an R-module. INTRODUCTORY NOTES ON MODULES 3 That addition in M is commutative actually follows from other axioms for R-modules: distribu- tivity both ways on (1 + 1)(m + m0) shows m + m + m0 + m0 = m + m0 + m + m0, and canceling the leftmost m's and rightmost m0's (addition on M is a group law) gives m + m0 = m0 + m for all m and m0 in M. We will show how many concepts from linear algebra (linear dependence/independence, basis, linear transformation, subspace) can be formulated for modules. When the scalars are not a field, we encounter genuinely new phenomena. For example, the intuitive idea of linear independence in a vector space as meaning no vector is a linear combination of the others is the wrong way to think about linear independence in modules. As another example, in a vector space each spanning set can be refined to a basis, but a module with a finite generating set does not have to have a basis (this is related to nonprincipal ideals). In the last section, we'll see how the concept of a module gives a new way to think about a question purely about matrices acting on vector spaces: for a field F and matrices A and B in Matn(F ), deciding if A and B are conjugate in Matn(F ) is the same as deciding if two F [T ]-module structures on F n (each one uses A or B) are isomorphic. 2. Basic definitions In linear algebra the concepts of linear combination, linear transformation, isomorphism, sub- space, and quotient space all make sense when the coefficients are in a ring, not just a field, so they can all be adapted to the setting of modules with no real changes. Definition 2.1. In an R-module M, an R-linear combination of elements m1; : : : ; mk 2 M is an element of M having the form r1m1 + ··· + rkmk where the ri's are in R. If every element of M is a linear combination of m1; : : : ; mk, we call fm1; : : : ; mkg a spanning set or generating set of M or say the mi's span (or generate) M. Linear combinations are the basic way to create new elements of a module from old ones, just as in linear algebra in Rn. For instance, an ideal (a; b; c) = Ra + Rb + Rc in R is nothing other than the set of R-linear combinations of a, b, and c in R. Example 2.2. We can view the ideal I = (1 + 2i) in Z[i] as both a Z[i]-module and as a Z-module in a natural way. As a Z[i]-module, we can get everywhere in I from 1+2i: I = Z[i](1+2i). As a Z- module, we can get everywhere in I from 1+2i and i(1+2i) = −2+i since I = Z(1+2i)+Zi(1+2i) = Z(1 + 2i) + Z(−2 + i). Mostly we will be interested in modules with finite spanning sets, but there are some important examples of modules that require infinite spanning sets. So let's give the general definition of spanning set, allowing infinite ones. 4 KEITH CONRAD Definition 2.3. A spanning set of an R-module M is a subset fmigi2I of M such that every m 2 M is a finite R-linear combination of the mi's: X m = rimi; i2I where ri 2 R for all i and ri = 0 for all but finitely many i. Notice in this definition that we require each linear combination of the mi's to have only finitely many nonzero coefficients. (Of course, if there are only finitely many mi's to begin with then this is no constraint at all.) In analysis, vector spaces may be equipped with a topology and we can talk about truly infinite linear combinations using a notion of convergence.
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